What is the integral from 0 to 4-y^2 of (2-z) ?

The integral from 0 to 4-y^2 of (2-z) is 2(-y^2+4)-((-y^2+4)^2)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4-y^2 of (2-z)

\int_{0}^{4 - y^{2}} (2 - z) d z

2 (- y^{2} + 4) - \frac{( - y ^{2} + 4 ) ^{2}}{2}

Solution steps

\int_{0}^{4 - y^{2}} 2 - z d z

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{4 - y^{2}} 2 d z - \int_{0}^{4 - y^{2}} z d z

\int_{0}^{4 - y^{2}} 2 d z = 2 (- y^{2} + 4)

\int_{0}^{4 - y^{2}} z d z = \frac{( - y ^{2} + 4 ) ^{2}}{2}

= 2 (- y^{2} + 4) - \frac{( - y ^{2} + 4 ) ^{2}}{2}

\int_{0}^{2} (x + 1)^{- 1} d x

\int_{0}^{5 - 2 x - \frac{1}{2} y} 6 z^{2} d z

\int_{1}^{4} ln (x^{2}) d x

\int_{- 1}^{1} x^{4} - x^{2} d x

\int_{- \frac{3}{2}}^{\frac{1}{2}} (x^{3} - 4 x) d x

What is the integral from 0 to 4-y^2 of (2-z) ?
The integral from 0 to 4-y^2 of (2-z) is 2(-y^2+4)-((-y^2+4)^2)/2